hdu 5059

Problem Description

As you know, when you want to hack someone's program, you must submit your test data. However sometimes you will submit invalid data, so we need a data checker to check your data. Now small W has prepared a problem for BC, but he is too busy to write the data checker. Please help him to write a data check which judges whether the input is an integer ranged from a to b (inclusive).
Note: a string represents a valid integer when it follows below rules.
1. When it represents a non-negative integer, it contains only digits without leading zeros.
2. When it represents a negative integer, it contains exact one negative sign ('-') followed by digits without leading zeros and there are no characters before '-'.
3. Otherwise it is not a valid integer.

 

 

Input

Multi test cases (about 100), every case occupies two lines, the first line contain a string which represents the input string, then second line contains a and b separated by space. Process to the end of file.

Length of string is no more than 100.
The string may contain any characters other than '\n','\r'.
-1000000000

≤a≤b≤1000000000

 

 

Output

For each case output "YES" (without quote) when the string is an integer ranged from a to b, otherwise output "NO" (without quote).

 

 

Sample Input

10

-100 100

1a0

-100 100

 

 

Sample Output

YES

NO

题意：第一行给出一个字符串 判断这个字符串是否合法 如果合法是否在区间【a，b】之内（包含a，b）   坑多。。。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e2+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;

typedef long long LL;

int main ()
{
    char s[N];
    LL a, b, c;
    int len, flag, i, mark, legal;

    while (gets(s) != 0)
    {
        scanf("%lld %lld%*c", &a, &b);
        len = strlen(s);
        mark = 0;///标记负号
        legal = 1;///是否合法
        c = 0;///保存字符串的值
        flag = 0;///0表示在范围内 1表示不在
        if (s[0] == '-') mark = 1;
        for (i = 0; i < len; i++)
        {
            if ((s[i] == '-' && i == 0) || (s[i] >= '0' && s[i] <= '9')) continue;
            /**<除了‘-’ 剩下的都必须是数字且不含有前导0的才是合法的串*/
            legal = 0;
            break;
        }

        i = 0;
        if (mark) i++;
        if (s[i] == '0') legal = 0;

        if (len > 15) legal = 0;///串太长超出范围

        if (legal)///只处理合法的串
        {
            i = 0;
            if (mark) i++;///排除‘-’的影响

            for ( ; i < len; i++)
                c = c*10+(s[i]-'0');

            if (mark) c = -c;///有‘-’ 取相反数

            if (c >= a && c <= b) flag = 1;
        }

        if (len == 1 && s[0] == '0' && 0 >= a && 0 <= b) flag = 1;///字符串只有一个‘0’如果在区间内也合法
        if (s[0] == '-' && len == 1) flag = 0;///只有‘-’ 也是不合法的
        if (len == 0) flag = 0;///空串是不合法的！！！！！

        if (flag == 0) printf("NO\n");
        else printf("YES\n");
    }

    return 0;
}